**Tree:-** A connected graph without any circuit is called a Tree. In other words, a tree is an undirected graph G that satisfies any of the following equivalent conditions:

- Any two vertices in G can be connected by a unique simple path.
- G is acyclic, and a simple cycle is formed if any edge is added to G.
- G is connected and has no cycles.
- G is connected but would become disconnected if any single edge is removed from G.
- G is connected and the 3-vertex complete graph K3 is not a minor of G.

**For Example**:

- This Graph is a Tree:

- This Graph is not a Tree:

Some theorems related to trees are:

**Theorem 1**: Prove that for a tree (T), there is one and only one path between every pair of vertices in a tree.**Proof:**Since tree (T) is a connected graph, there exist at least one path between every pair of vertices in a tree (T). Now, suppose between two vertices a and b of the tree (T) there exist two paths. The union of these two paths will contain a circuit and tree (T) cannot be a tree. Hence the above statement is proved.**Theorem 2:**If in a graph G there is one and only one path between every pair of vertices than graph G is a tree.**Proof:**There is the existence of a path between every pair of vertices so we assume that graph G is connected. A circuit in a graph implies that there is at least one pair of vertices a and b, such that there are two distinct paths between a and b. Since G has one and only one path between every pair of vertices. G cannot have any circuit. Hence graph G is a tree.

**Theorem 3:**Prove that a tree with n vertices has (n-1) edges.**Proof:**Let n be the number of vertices in a tree (T).

If n=1, then the number of edges=0.

If n=2 then the number of edges=1.

If n=3 then the number of edges=2.Hence, the statement (or result) is true for n=1, 2, 3.

Let the statement be true for n=m. Now we want to prove that it is true for n=m+1.

Let be the edge connecting vertices say Vi and Vj. Since G is a tree, then there exists only one path between vertices Vi and Vj. Hence if we delete edge e it will be disconnected graph into two components G1 and G2 say. These components have less than m+1 vertices and there is no circuit and hence each component G1 and G2 have m1 and m2 vertices.

Now, the total no. of edges = (m1-1) + (m2-1) +1 = (m1+m2)-1 = m+1-1 = m.

Hence for

**n=m+1**vertices there are m edges in a tree (T). By the mathematical induction the graph exactly has n-1 edges.**Theorem 4:**Prove that any connected graph G with n vertices and (n-1) edges is a tree.**Proof:**We know that the minimum number of edges required to make a graph of n vertices connected is (n-1) edges. We can observe that removal of one edge from the graph G will make it disconnected. Thus a connected graph of n vertices and (n-1) edges cannot have a circuit. Hence a graph G is a tree.**Theorem 5:**Prove that a graph with n vertices,**(n-1)**edges and no circuit is a connected graph.**Proof:**Let the graph G is disconnected then there exist at least two components G1 and G2 say. Each of the component is circuit-less as G is circuit-less. Now to make a graph G connected we need to add one edge e between the vertices Vi and Vj, where Vi is the vertex of G1 and Vj is the vertex of component G2.

Now the number of edges in**G = (n – 1)+1 =n.**Now, G is connected graph and circuit-less with n vertices and n edges, which is impossible because the connected circuit-less graph is a tree and tree with n vertices has

**(n-1)**edges. So the graph G with n vertices,**(n-1)**edges and without circuit is connected. Hence the given statement is proved.

**Theorem 6:**A graph G is a tree if and only if it is minimally connected.**Proof:**Let the graph G is minimally connected, i.e; removal of one edge make it disconnected. Therefore, there is no circuit. Hence graph G is a tree.

Conversely, let the graph G is a tree i.e; there exists one and only one path between every pair of vertices and we know that removal of one edge from the path makes the graph disconnected. Hence graph G is minimally connected.**Theorem 7:**Every tree with at-least two vertices has at-least two pendant vertices.**Proof:**Let the number of vertices in a given tree T is n and n>=2. Therefore the number of edges in a tree T=n-1 using above theorems.summation of (deg(Vi)) = 2*e = 2*(n-1) =2n-2

The degree sum is to be divided among n vertices. Since a tree T is a connected graph, it cannot have a vertex of degree zero. Each vertex contributes at-least one to the above sum. Thus there must be at least two vertices of degree 1. Hence every tree with at-least two vertices have at-least two pendant vertices.

**Theorem 8:**Show that every tree has either one or two centres.**Proof:**We will use one observation that the maximum distance max d(v, w) from a given vertex v to any other vertex w occurs only when w is pendant vertex.Now, let T is a tree with n vertices (n>=2)

T must have at least two pendant vertices. Delete all pendant vertices from T, then resulting graph T’ is still a tree. Again delete pendant vertices from T’ so that resulting T” is still a tree with same centers.

Note that all vertices that T had as centers will still remain centers in T’–>T”–>T”’–>….

continue this process until the remaining tree has either one vertex or one edge. So in the end, if one vertex is there this implies tree T has one center. If one edge is there then tree T has two centers.

**Theorem 9:**Prove the maximum number of vertices at level ‘L’ in a binary tree is , where L>=0.**Proof:**The given theorem is proved with the help of mathematical induction. At level 0 (L=0), there is only one vertex at level (L=1), there is only vertices.Now we assume that statement is true for the level (L-1).

Therefore, maximum number of vertices on the level (L-1) is . Since we know that each vertex in a binary tree has the maximum of 2 vertices in next level, therefore the number of vertices on the level L is twice that of the level L-1.

Hence at level L, the number of vertices is:-

* = .

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